Integration reverses differentiation. We will practice “undoing” the power rule by finding an antiderivative.
Let \(f(x)\) be an unknown polynomial. Its derivative is known.
\[f'(x)={-15}x^{4}+{16}x^{3}-{18}x^{2}\]
To undo the power rule, each term’s power is increased by 1, and then the coefficient is divided by the new power. This gets us most of the way.
A (slightly annoying) complication occurs because adding any constant to \(f(x)\) would not change its derivative, since the derivative of a constant is 0. This means there are infinitely many possible antiderivatives with a constant of integration equaling any real number. If we know a single solution to \(y=f(x)\) (a single point on the curve), then the constant of integration can be determined. In this case, let’s enforce the following condition:
\[f(2)=-84\]
We know the integral can be expressed as a polynomial with 4 terms, where \(j>k>l\):
\[f(x)=Jx^{j}+Kx^{k}+Lx^l+C\]
Find the parameters.
\(J=\)
\(j=\)
\(K=\)
\(k=\)
\(L=\)
\(l=\)
\(C=\)
Solution
We were given the derivative.
\[f'(x)={-15}x^{4}+{16}x^{3}-{18}x^{2}\]
The degrees of \(f'(x)\) are 4, 3, 2. If we add 1 to each, we get the new degrees.
\[f(x)={J}x^{5}+{K}x^{4}+{L}x^{3}+C\]
Divide the coefficients of \(f'(x)\) by the new degrees.
Integration reverses differentiation. We will practice “undoing” the power rule by finding an antiderivative.
Let \(f(x)\) be an unknown polynomial. Its derivative is known.
\[f'(x)={-10}x^{4}-{32}x^{3}-{4}{}\]
To undo the power rule, each term’s power is increased by 1, and then the coefficient is divided by the new power. This gets us most of the way.
A (slightly annoying) complication occurs because adding any constant to \(f(x)\) would not change its derivative, since the derivative of a constant is 0. This means there are infinitely many possible antiderivatives with a constant of integration equaling any real number. If we know a single solution to \(y=f(x)\) (a single point on the curve), then the constant of integration can be determined. In this case, let’s enforce the following condition:
\[f(-4)=11\]
We know the integral can be expressed as a polynomial with 4 terms, where \(j>k>l\):
\[f(x)=Jx^{j}+Kx^{k}+Lx^l+C\]
Find the parameters.
\(J=\)
\(j=\)
\(K=\)
\(k=\)
\(L=\)
\(l=\)
\(C=\)
Solution
We were given the derivative.
\[f'(x)={-10}x^{4}-{32}x^{3}-{4}{}\]
The degrees of \(f'(x)\) are 4, 3, 0. If we add 1 to each, we get the new degrees.
\[f(x)={J}x^{5}+{K}x^{4}+{L}x^{1}+C\]
Divide the coefficients of \(f'(x)\) by the new degrees.
Integration reverses differentiation. We will practice “undoing” the power rule by finding an antiderivative.
Let \(f(x)\) be an unknown polynomial. Its derivative is known.
\[f'(x)={-12}x^{3}+{24}x^{2}-{18}x^{}\]
To undo the power rule, each term’s power is increased by 1, and then the coefficient is divided by the new power. This gets us most of the way.
A (slightly annoying) complication occurs because adding any constant to \(f(x)\) would not change its derivative, since the derivative of a constant is 0. This means there are infinitely many possible antiderivatives with a constant of integration equaling any real number. If we know a single solution to \(y=f(x)\) (a single point on the curve), then the constant of integration can be determined. In this case, let’s enforce the following condition:
\[f(2)=-26\]
We know the integral can be expressed as a polynomial with 4 terms, where \(j>k>l\):
\[f(x)=Jx^{j}+Kx^{k}+Lx^l+C\]
Find the parameters.
\(J=\)
\(j=\)
\(K=\)
\(k=\)
\(L=\)
\(l=\)
\(C=\)
Solution
We were given the derivative.
\[f'(x)={-12}x^{3}+{24}x^{2}-{18}x^{}\]
The degrees of \(f'(x)\) are 3, 2, 1. If we add 1 to each, we get the new degrees.
\[f(x)={J}x^{4}+{K}x^{3}+{L}x^{2}+C\]
Divide the coefficients of \(f'(x)\) by the new degrees.
Integration reverses differentiation. We will practice “undoing” the power rule by finding an antiderivative.
Let \(f(x)\) be an unknown polynomial. Its derivative is known.
\[f'(x)={-35}x^{6}-{36}x^{5}+{35}x^{4}\]
To undo the power rule, each term’s power is increased by 1, and then the coefficient is divided by the new power. This gets us most of the way.
A (slightly annoying) complication occurs because adding any constant to \(f(x)\) would not change its derivative, since the derivative of a constant is 0. This means there are infinitely many possible antiderivatives with a constant of integration equaling any real number. If we know a single solution to \(y=f(x)\) (a single point on the curve), then the constant of integration can be determined. In this case, let’s enforce the following condition:
\[f(-2)=39\]
We know the integral can be expressed as a polynomial with 4 terms, where \(j>k>l\):
\[f(x)=Jx^{j}+Kx^{k}+Lx^l+C\]
Find the parameters.
\(J=\)
\(j=\)
\(K=\)
\(k=\)
\(L=\)
\(l=\)
\(C=\)
Solution
We were given the derivative.
\[f'(x)={-35}x^{6}-{36}x^{5}+{35}x^{4}\]
The degrees of \(f'(x)\) are 6, 5, 4. If we add 1 to each, we get the new degrees.
\[f(x)={J}x^{7}+{K}x^{6}+{L}x^{5}+C\]
Divide the coefficients of \(f'(x)\) by the new degrees.
Integration reverses differentiation. We will practice “undoing” the power rule by finding an antiderivative.
Let \(f(x)\) be an unknown polynomial. Its derivative is known.
\[f'(x)={20}x^{4}-{12}x^{3}-{27}x^{2}\]
To undo the power rule, each term’s power is increased by 1, and then the coefficient is divided by the new power. This gets us most of the way.
A (slightly annoying) complication occurs because adding any constant to \(f(x)\) would not change its derivative, since the derivative of a constant is 0. This means there are infinitely many possible antiderivatives with a constant of integration equaling any real number. If we know a single solution to \(y=f(x)\) (a single point on the curve), then the constant of integration can be determined. In this case, let’s enforce the following condition:
\[f(2)=2\]
We know the integral can be expressed as a polynomial with 4 terms, where \(j>k>l\):
\[f(x)=Jx^{j}+Kx^{k}+Lx^l+C\]
Find the parameters.
\(J=\)
\(j=\)
\(K=\)
\(k=\)
\(L=\)
\(l=\)
\(C=\)
Solution
We were given the derivative.
\[f'(x)={20}x^{4}-{12}x^{3}-{27}x^{2}\]
The degrees of \(f'(x)\) are 4, 3, 2. If we add 1 to each, we get the new degrees.
\[f(x)={J}x^{5}+{K}x^{4}+{L}x^{3}+C\]
Divide the coefficients of \(f'(x)\) by the new degrees.
A definite integral can be used to express the area under a curve. For example, say we wanted to find the area under \(f(x)=x^{}-x^{7}\) between \(x=0.43\) and \(x=0.81\). Let’s draw a picture.
The area equals the definite integral of the function on interval [0.43, 0.81].
A definite integral can be used to express the area under a curve. For example, say we wanted to find the area under \(f(x)=x^{3}-x^{9}\) between \(x=0.51\) and \(x=0.8\). Let’s draw a picture.
The area equals the definite integral of the function on interval [0.51, 0.8].
A definite integral can be used to express the area under a curve. For example, say we wanted to find the area under \(f(x)=x^{}-x^{6}\) between \(x=0.49\) and \(x=0.87\). Let’s draw a picture.
The area equals the definite integral of the function on interval [0.49, 0.87].
A definite integral can be used to express the area under a curve. For example, say we wanted to find the area under \(f(x)=x^{}-x^{5}\) between \(x=0.54\) and \(x=0.8\). Let’s draw a picture.
The area equals the definite integral of the function on interval [0.54, 0.8].
A definite integral can be used to express the area under a curve. For example, say we wanted to find the area under \(f(x)=x^{3}-x^{7}\) between \(x=0.56\) and \(x=0.85\). Let’s draw a picture.
The area equals the definite integral of the function on interval [0.56, 0.85].
A particle moves up and down along a 1-dimensional path. It’s acceleration (\(a(t)\) in \(\mathrm{\frac{m}{s^2}}\)) is a function of time (\(t\) in seconds).
\[a(t)=-48t+12\]
The particle’s velocity is also a function of time. Also, \(v(0)=4\).
\[v(t) ~=~ 4 + \int_{0}^{t}a(t)\,dt\]
The particle’s position is also a function of time. Also, \(x(0)=4\).
\[x(t) ~=~ 4 + \int_{0}^{t}v(t)\,dt\]
Evaluate position, velocity, and acceleration at \(t=0\), \(t=1\), and \(t=2\).
t
y(t)
v(t)
a(t)
0
1
2
Solution
We were given acceleration as a function of time.
\[a(t)=-48t+12\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[v(t)=-24t^2+12t+4\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[y(t)=-8t^3+6t^2+4t+4\]
t
y(t)
v(t)
a(t)
0
4
4
12
1
6
-8
-36
2
-28
-68
-84
Question
A particle moves up and down along a 1-dimensional path. It’s acceleration (\(a(t)\) in \(\mathrm{\frac{m}{s^2}}\)) is a function of time (\(t\) in seconds).
\[a(t)=-36t+10\]
The particle’s velocity is also a function of time. Also, \(v(0)=8\).
\[v(t) ~=~ 8 + \int_{0}^{t}a(t)\,dt\]
The particle’s position is also a function of time. Also, \(x(0)=2\).
\[x(t) ~=~ 2 + \int_{0}^{t}v(t)\,dt\]
Evaluate position, velocity, and acceleration at \(t=0\), \(t=1\), and \(t=2\).
t
y(t)
v(t)
a(t)
0
1
2
Solution
We were given acceleration as a function of time.
\[a(t)=-36t+10\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[v(t)=-18t^2+10t+8\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[y(t)=-6t^3+5t^2+8t+2\]
t
y(t)
v(t)
a(t)
0
2
8
10
1
9
0
-26
2
-10
-44
-62
Question
A particle moves up and down along a 1-dimensional path. It’s acceleration (\(a(t)\) in \(\mathrm{\frac{m}{s^2}}\)) is a function of time (\(t\) in seconds).
\[a(t)=-48t+20\]
The particle’s velocity is also a function of time. Also, \(v(0)=6\).
\[v(t) ~=~ 6 + \int_{0}^{t}a(t)\,dt\]
The particle’s position is also a function of time. Also, \(x(0)=-1\).
\[x(t) ~=~ -1 + \int_{0}^{t}v(t)\,dt\]
Evaluate position, velocity, and acceleration at \(t=0\), \(t=1\), and \(t=2\).
t
y(t)
v(t)
a(t)
0
1
2
Solution
We were given acceleration as a function of time.
\[a(t)=-48t+20\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[v(t)=-24t^2+20t+6\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[y(t)=-8t^3+10t^2+6t-1\]
t
y(t)
v(t)
a(t)
0
-1
6
20
1
7
2
-28
2
-13
-50
-76
Question
A particle moves up and down along a 1-dimensional path. It’s acceleration (\(a(t)\) in \(\mathrm{\frac{m}{s^2}}\)) is a function of time (\(t\) in seconds).
\[a(t)=-36t+4\]
The particle’s velocity is also a function of time. Also, \(v(0)=8\).
\[v(t) ~=~ 8 + \int_{0}^{t}a(t)\,dt\]
The particle’s position is also a function of time. Also, \(x(0)=10\).
\[x(t) ~=~ 10 + \int_{0}^{t}v(t)\,dt\]
Evaluate position, velocity, and acceleration at \(t=0\), \(t=1\), and \(t=2\).
t
y(t)
v(t)
a(t)
0
1
2
Solution
We were given acceleration as a function of time.
\[a(t)=-36t+4\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[v(t)=-18t^2+4t+8\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[y(t)=-6t^3+2t^2+8t+10\]
t
y(t)
v(t)
a(t)
0
10
8
4
1
14
-6
-32
2
-14
-56
-68
Question
A particle moves up and down along a 1-dimensional path. It’s acceleration (\(a(t)\) in \(\mathrm{\frac{m}{s^2}}\)) is a function of time (\(t\) in seconds).
\[a(t)=-54t+14\]
The particle’s velocity is also a function of time. Also, \(v(0)=9\).
\[v(t) ~=~ 9 + \int_{0}^{t}a(t)\,dt\]
The particle’s position is also a function of time. Also, \(x(0)=2\).
\[x(t) ~=~ 2 + \int_{0}^{t}v(t)\,dt\]
Evaluate position, velocity, and acceleration at \(t=0\), \(t=1\), and \(t=2\).
t
y(t)
v(t)
a(t)
0
1
2
Solution
We were given acceleration as a function of time.
\[a(t)=-54t+14\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[v(t)=-27t^2+14t+9\]
To find the antiderivative, add 1 to each power and then divide each coefficient by the new power. Then, add the constant of integration to match the given initial value.
\[y(t)=-9t^3+7t^2+9t+2\]
t
y(t)
v(t)
a(t)
0
2
9
14
1
9
-4
-40
2
-24
-71
-94
Question
A 3D shape’s base is the 2D region enclosed by \(y=0\) and \(y=f(x)=x^{0.8}-x^{5.5}\) on the \(xy\) plane. For every plane perpendicular to the \(x\) axis, with \(0<x<1\), there is a square cross section.
I have attempted to draw this below.
And here is a spinning animation. Does that help?
Find the volume of the shape; the tolerance is \(\pm 0.001\) cubic units. You might find this video helpful. The general topic is “volume from cross sections”; you might try a variety of resources by searching this phrase in your favorite search engine.
Solution
The volume can be found by accumulating the volumes of many “slabs”, where each slab is a square with some infinitesimal thickness.
The volume of each slab depends on \(x\). The width and height are each \(f(x)\) and the thickness is an infinitesimal\(dx\).
So, we can find the infinitesimal volume as a function of \(x\).
\[dV = (f(x))^2 \,dx\]
\[dV = (x^{0.8}-x^{5.5})^2 \,dx\]
We need to sum up all the differential volumes from \(x=0\) to \(x=1\).
A 3D shape’s base is the 2D region enclosed by \(y=0\) and \(y=f(x)=x^{0.5}-x^{2.4}\) on the \(xy\) plane. For every plane perpendicular to the \(x\) axis, with \(0<x<1\), there is a square cross section.
I have attempted to draw this below.
And here is a spinning animation. Does that help?
Find the volume of the shape; the tolerance is \(\pm 0.001\) cubic units. You might find this video helpful. The general topic is “volume from cross sections”; you might try a variety of resources by searching this phrase in your favorite search engine.
Solution
The volume can be found by accumulating the volumes of many “slabs”, where each slab is a square with some infinitesimal thickness.
The volume of each slab depends on \(x\). The width and height are each \(f(x)\) and the thickness is an infinitesimal\(dx\).
So, we can find the infinitesimal volume as a function of \(x\).
\[dV = (f(x))^2 \,dx\]
\[dV = (x^{0.5}-x^{2.4})^2 \,dx\]
We need to sum up all the differential volumes from \(x=0\) to \(x=1\).
A 3D shape’s base is the 2D region enclosed by \(y=0\) and \(y=f(x)=x^{0.7}-x^{8.6}\) on the \(xy\) plane. For every plane perpendicular to the \(x\) axis, with \(0<x<1\), there is a square cross section.
I have attempted to draw this below.
And here is a spinning animation. Does that help?
Find the volume of the shape; the tolerance is \(\pm 0.001\) cubic units. You might find this video helpful. The general topic is “volume from cross sections”; you might try a variety of resources by searching this phrase in your favorite search engine.
Solution
The volume can be found by accumulating the volumes of many “slabs”, where each slab is a square with some infinitesimal thickness.
The volume of each slab depends on \(x\). The width and height are each \(f(x)\) and the thickness is an infinitesimal\(dx\).
So, we can find the infinitesimal volume as a function of \(x\).
\[dV = (f(x))^2 \,dx\]
\[dV = (x^{0.7}-x^{8.6})^2 \,dx\]
We need to sum up all the differential volumes from \(x=0\) to \(x=1\).
A 3D shape’s base is the 2D region enclosed by \(y=0\) and \(y=f(x)=x^{1.7}-x^{9.3}\) on the \(xy\) plane. For every plane perpendicular to the \(x\) axis, with \(0<x<1\), there is a square cross section.
I have attempted to draw this below.
And here is a spinning animation. Does that help?
Find the volume of the shape; the tolerance is \(\pm 0.001\) cubic units. You might find this video helpful. The general topic is “volume from cross sections”; you might try a variety of resources by searching this phrase in your favorite search engine.
Solution
The volume can be found by accumulating the volumes of many “slabs”, where each slab is a square with some infinitesimal thickness.
The volume of each slab depends on \(x\). The width and height are each \(f(x)\) and the thickness is an infinitesimal\(dx\).
So, we can find the infinitesimal volume as a function of \(x\).
\[dV = (f(x))^2 \,dx\]
\[dV = (x^{1.7}-x^{9.3})^2 \,dx\]
We need to sum up all the differential volumes from \(x=0\) to \(x=1\).
A 3D shape’s base is the 2D region enclosed by \(y=0\) and \(y=f(x)=x^{1.1}-x^{3.6}\) on the \(xy\) plane. For every plane perpendicular to the \(x\) axis, with \(0<x<1\), there is a square cross section.
I have attempted to draw this below.
And here is a spinning animation. Does that help?
Find the volume of the shape; the tolerance is \(\pm 0.001\) cubic units. You might find this video helpful. The general topic is “volume from cross sections”; you might try a variety of resources by searching this phrase in your favorite search engine.
Solution
The volume can be found by accumulating the volumes of many “slabs”, where each slab is a square with some infinitesimal thickness.
The volume of each slab depends on \(x\). The width and height are each \(f(x)\) and the thickness is an infinitesimal\(dx\).
So, we can find the infinitesimal volume as a function of \(x\).
\[dV = (f(x))^2 \,dx\]
\[dV = (x^{1.1}-x^{3.6})^2 \,dx\]
We need to sum up all the differential volumes from \(x=0\) to \(x=1\).
A 3D shape is produced with elliptical cross sections. As \(x\) progresses from along the interval \([0,1]\), the cross section perpendicular to the \(x\) axis will be an ellipse with axes (and radii) parallel to the \(y\) and \(z\) axes. The (maximum and minimum) radii are power functions with respect to \(x\).
\[r_1(x) ~=~ x^{1.45}\]\[r_2(x) ~=~ x^{0.38}\]
Find the volume of the shape. The tolerance is \(\pm 0.01\) cubic units.
As a hint, I’ll remind you that the area of an ellipse is found by multiplying the radii by each other and pi: \(A=\pi\cdot r_1 \cdot r_2\).
Solution
Imagine breaking the 3D shape into a bunch of elliptical slices with infinitesimal thickness \(dx\).
Each elliptical slab has an infinitesimal volume.
\[dV = \pi r_1 r_2 \, dx\]
\[dV = \pi x^{1.45} x^{0.38} \, dx\]
\[dV = \pi x^{1.83} \, dx\]
To get the total volume, we need to Sum up all the infinitesimal volumes as \(x\) varies from 0 to 1.
\[V = \int_{x=0}^{x=1} dV\]
\[V = \int_{0}^{1} \pi x^{1.83} \, dx\]
Find antiderivitive \(g(x)\) so that \(g'(x)=\pi x^{1.83}\) and \(g(0)=0\) (so the constant of integration is 0).
\[g(x) = \frac{\pi}{2.83} x^{2.83}\]
The total volume can be expressed in terms of function \(g\).
A 3D shape is produced with elliptical cross sections. As \(x\) progresses from along the interval \([0,1]\), the cross section perpendicular to the \(x\) axis will be an ellipse with axes (and radii) parallel to the \(y\) and \(z\) axes. The (maximum and minimum) radii are power functions with respect to \(x\).
\[r_1(x) ~=~ x^{1.03}\]\[r_2(x) ~=~ x^{2.52}\]
Find the volume of the shape. The tolerance is \(\pm 0.01\) cubic units.
As a hint, I’ll remind you that the area of an ellipse is found by multiplying the radii by each other and pi: \(A=\pi\cdot r_1 \cdot r_2\).
Solution
Imagine breaking the 3D shape into a bunch of elliptical slices with infinitesimal thickness \(dx\).
Each elliptical slab has an infinitesimal volume.
\[dV = \pi r_1 r_2 \, dx\]
\[dV = \pi x^{1.03} x^{2.52} \, dx\]
\[dV = \pi x^{3.55} \, dx\]
To get the total volume, we need to Sum up all the infinitesimal volumes as \(x\) varies from 0 to 1.
\[V = \int_{x=0}^{x=1} dV\]
\[V = \int_{0}^{1} \pi x^{3.55} \, dx\]
Find antiderivitive \(g(x)\) so that \(g'(x)=\pi x^{3.55}\) and \(g(0)=0\) (so the constant of integration is 0).
\[g(x) = \frac{\pi}{4.55} x^{4.55}\]
The total volume can be expressed in terms of function \(g\).
A 3D shape is produced with elliptical cross sections. As \(x\) progresses from along the interval \([0,1]\), the cross section perpendicular to the \(x\) axis will be an ellipse with axes (and radii) parallel to the \(y\) and \(z\) axes. The (maximum and minimum) radii are power functions with respect to \(x\).
\[r_1(x) ~=~ x^{0.67}\]\[r_2(x) ~=~ x^{1}\]
Find the volume of the shape. The tolerance is \(\pm 0.01\) cubic units.
As a hint, I’ll remind you that the area of an ellipse is found by multiplying the radii by each other and pi: \(A=\pi\cdot r_1 \cdot r_2\).
Solution
Imagine breaking the 3D shape into a bunch of elliptical slices with infinitesimal thickness \(dx\).
Each elliptical slab has an infinitesimal volume.
\[dV = \pi r_1 r_2 \, dx\]
\[dV = \pi x^{0.67} x^{1} \, dx\]
\[dV = \pi x^{1.67} \, dx\]
To get the total volume, we need to Sum up all the infinitesimal volumes as \(x\) varies from 0 to 1.
\[V = \int_{x=0}^{x=1} dV\]
\[V = \int_{0}^{1} \pi x^{1.67} \, dx\]
Find antiderivitive \(g(x)\) so that \(g'(x)=\pi x^{1.67}\) and \(g(0)=0\) (so the constant of integration is 0).
\[g(x) = \frac{\pi}{2.67} x^{2.67}\]
The total volume can be expressed in terms of function \(g\).
A 3D shape is produced with elliptical cross sections. As \(x\) progresses from along the interval \([0,1]\), the cross section perpendicular to the \(x\) axis will be an ellipse with axes (and radii) parallel to the \(y\) and \(z\) axes. The (maximum and minimum) radii are power functions with respect to \(x\).
\[r_1(x) ~=~ x^{2.11}\]\[r_2(x) ~=~ x^{1.58}\]
Find the volume of the shape. The tolerance is \(\pm 0.01\) cubic units.
As a hint, I’ll remind you that the area of an ellipse is found by multiplying the radii by each other and pi: \(A=\pi\cdot r_1 \cdot r_2\).
Solution
Imagine breaking the 3D shape into a bunch of elliptical slices with infinitesimal thickness \(dx\).
Each elliptical slab has an infinitesimal volume.
\[dV = \pi r_1 r_2 \, dx\]
\[dV = \pi x^{2.11} x^{1.58} \, dx\]
\[dV = \pi x^{3.69} \, dx\]
To get the total volume, we need to Sum up all the infinitesimal volumes as \(x\) varies from 0 to 1.
\[V = \int_{x=0}^{x=1} dV\]
\[V = \int_{0}^{1} \pi x^{3.69} \, dx\]
Find antiderivitive \(g(x)\) so that \(g'(x)=\pi x^{3.69}\) and \(g(0)=0\) (so the constant of integration is 0).
\[g(x) = \frac{\pi}{4.69} x^{4.69}\]
The total volume can be expressed in terms of function \(g\).
A 3D shape is produced with elliptical cross sections. As \(x\) progresses from along the interval \([0,1]\), the cross section perpendicular to the \(x\) axis will be an ellipse with axes (and radii) parallel to the \(y\) and \(z\) axes. The (maximum and minimum) radii are power functions with respect to \(x\).
\[r_1(x) ~=~ x^{1.86}\]\[r_2(x) ~=~ x^{1.64}\]
Find the volume of the shape. The tolerance is \(\pm 0.01\) cubic units.
As a hint, I’ll remind you that the area of an ellipse is found by multiplying the radii by each other and pi: \(A=\pi\cdot r_1 \cdot r_2\).
Solution
Imagine breaking the 3D shape into a bunch of elliptical slices with infinitesimal thickness \(dx\).
Each elliptical slab has an infinitesimal volume.
\[dV = \pi r_1 r_2 \, dx\]
\[dV = \pi x^{1.86} x^{1.64} \, dx\]
\[dV = \pi x^{3.5} \, dx\]
To get the total volume, we need to Sum up all the infinitesimal volumes as \(x\) varies from 0 to 1.
\[V = \int_{x=0}^{x=1} dV\]
\[V = \int_{0}^{1} \pi x^{3.5} \, dx\]
Find antiderivitive \(g(x)\) so that \(g'(x)=\pi x^{3.5}\) and \(g(0)=0\) (so the constant of integration is 0).
\[g(x) = \frac{\pi}{4.5} x^{4.5}\]
The total volume can be expressed in terms of function \(g\).
A bowl is designed as the revolution of the region between 3 curves:
\[z=4x-44\]
\[z=\frac{4}{144}x^2\]
\[z=0\]
The revolution occurs around the \(z\) axis. The bowl will be made of wood.
Notice the point of intersection is \((12, 4)\). Below is a wireframe animation of the bowl.
Find the volume of the wood composing the bowl. I’d recommend using the washer method. The tolerance is \(\pm 1\) cubic units.
Solution
We will use the washer method. Notice we can approximate the shape with a bunch of thin washers.
Each washer has two radii (inner and outer) and an infinitesimal thickness.
The infinitesimal volume for each washer can be expressed in terms of \(r\), \(R\), and \(dz\).
\[dV ~=~ \pi \left(R^2 - r^2\right) \, dz\]
We need to express \(R\) and \(r\) as functions of \(z\). Notice that \(R\) is equivalent to \(x\) in the linear equation and \(r\) is equivalent to \(x\) in the quadratic equation. Let’s start by finding \(R\) in terms of \(z\).
\[z = 4R-44\]
Solve for \(R\).
\[R = \frac{z+44}{4}\]
Now, let’s get \(r\) in terms of \(z\).
\[z = \frac{4}{144}r^2\]
Solve for \(r\).
\[r = \sqrt{\frac{144z}{4}}\]
To get the volume, we need to sum up a bunch of washer volumes as \(z\) progresses from 0 to 4. Notice, \(4\) comes from the intersection at (12, 4).
A bowl is designed as the revolution of the region between 3 curves:
\[z=4x-30\]
\[z=\frac{6}{81}x^2\]
\[z=0\]
The revolution occurs around the \(z\) axis. The bowl will be made of wood.
Notice the point of intersection is \((9, 6)\). Below is a wireframe animation of the bowl.
Find the volume of the wood composing the bowl. I’d recommend using the washer method. The tolerance is \(\pm 1\) cubic units.
Solution
We will use the washer method. Notice we can approximate the shape with a bunch of thin washers.
Each washer has two radii (inner and outer) and an infinitesimal thickness.
The infinitesimal volume for each washer can be expressed in terms of \(r\), \(R\), and \(dz\).
\[dV ~=~ \pi \left(R^2 - r^2\right) \, dz\]
We need to express \(R\) and \(r\) as functions of \(z\). Notice that \(R\) is equivalent to \(x\) in the linear equation and \(r\) is equivalent to \(x\) in the quadratic equation. Let’s start by finding \(R\) in terms of \(z\).
\[z = 4R-30\]
Solve for \(R\).
\[R = \frac{z+30}{4}\]
Now, let’s get \(r\) in terms of \(z\).
\[z = \frac{6}{81}r^2\]
Solve for \(r\).
\[r = \sqrt{\frac{81z}{6}}\]
To get the volume, we need to sum up a bunch of washer volumes as \(z\) progresses from 0 to 6. Notice, \(6\) comes from the intersection at (9, 6).
A bowl is designed as the revolution of the region between 3 curves:
\[z=3x-25\]
\[z=\frac{8}{121}x^2\]
\[z=0\]
The revolution occurs around the \(z\) axis. The bowl will be made of wood.
Notice the point of intersection is \((11, 8)\). Below is a wireframe animation of the bowl.
Find the volume of the wood composing the bowl. I’d recommend using the washer method. The tolerance is \(\pm 1\) cubic units.
Solution
We will use the washer method. Notice we can approximate the shape with a bunch of thin washers.
Each washer has two radii (inner and outer) and an infinitesimal thickness.
The infinitesimal volume for each washer can be expressed in terms of \(r\), \(R\), and \(dz\).
\[dV ~=~ \pi \left(R^2 - r^2\right) \, dz\]
We need to express \(R\) and \(r\) as functions of \(z\). Notice that \(R\) is equivalent to \(x\) in the linear equation and \(r\) is equivalent to \(x\) in the quadratic equation. Let’s start by finding \(R\) in terms of \(z\).
\[z = 3R-25\]
Solve for \(R\).
\[R = \frac{z+25}{3}\]
Now, let’s get \(r\) in terms of \(z\).
\[z = \frac{8}{121}r^2\]
Solve for \(r\).
\[r = \sqrt{\frac{121z}{8}}\]
To get the volume, we need to sum up a bunch of washer volumes as \(z\) progresses from 0 to 8. Notice, \(8\) comes from the intersection at (11, 8).
A bowl is designed as the revolution of the region between 3 curves:
\[z=2x-11\]
\[z=\frac{9}{100}x^2\]
\[z=0\]
The revolution occurs around the \(z\) axis. The bowl will be made of wood.
Notice the point of intersection is \((10, 9)\). Below is a wireframe animation of the bowl.
Find the volume of the wood composing the bowl. I’d recommend using the washer method. The tolerance is \(\pm 1\) cubic units.
Solution
We will use the washer method. Notice we can approximate the shape with a bunch of thin washers.
Each washer has two radii (inner and outer) and an infinitesimal thickness.
The infinitesimal volume for each washer can be expressed in terms of \(r\), \(R\), and \(dz\).
\[dV ~=~ \pi \left(R^2 - r^2\right) \, dz\]
We need to express \(R\) and \(r\) as functions of \(z\). Notice that \(R\) is equivalent to \(x\) in the linear equation and \(r\) is equivalent to \(x\) in the quadratic equation. Let’s start by finding \(R\) in terms of \(z\).
\[z = 2R-11\]
Solve for \(R\).
\[R = \frac{z+11}{2}\]
Now, let’s get \(r\) in terms of \(z\).
\[z = \frac{9}{100}r^2\]
Solve for \(r\).
\[r = \sqrt{\frac{100z}{9}}\]
To get the volume, we need to sum up a bunch of washer volumes as \(z\) progresses from 0 to 9. Notice, \(9\) comes from the intersection at (10, 9).
A bowl is designed as the revolution of the region between 3 curves:
\[z=5x-61\]
\[z=\frac{4}{169}x^2\]
\[z=0\]
The revolution occurs around the \(z\) axis. The bowl will be made of wood.
Notice the point of intersection is \((13, 4)\). Below is a wireframe animation of the bowl.
Find the volume of the wood composing the bowl. I’d recommend using the washer method. The tolerance is \(\pm 1\) cubic units.
Solution
We will use the washer method. Notice we can approximate the shape with a bunch of thin washers.
Each washer has two radii (inner and outer) and an infinitesimal thickness.
The infinitesimal volume for each washer can be expressed in terms of \(r\), \(R\), and \(dz\).
\[dV ~=~ \pi \left(R^2 - r^2\right) \, dz\]
We need to express \(R\) and \(r\) as functions of \(z\). Notice that \(R\) is equivalent to \(x\) in the linear equation and \(r\) is equivalent to \(x\) in the quadratic equation. Let’s start by finding \(R\) in terms of \(z\).
\[z = 5R-61\]
Solve for \(R\).
\[R = \frac{z+61}{5}\]
Now, let’s get \(r\) in terms of \(z\).
\[z = \frac{4}{169}r^2\]
Solve for \(r\).
\[r = \sqrt{\frac{169z}{4}}\]
To get the volume, we need to sum up a bunch of washer volumes as \(z\) progresses from 0 to 4. Notice, \(4\) comes from the intersection at (13, 4).
\[dV ~=~ \pi r^2 h + 2\pi r h\,dr + \pi h \, (dr)^2-\pi r^2h\]
Notice two terms eliminate because they are opposites.
\[dV ~=~ 2\pi r h\,dr + \pi h \, (dr)^2\]
Remember, \(dr\) is an infinitesimal amount (\(dr\) is near 0), so the second term can be ignored. This leaves us with the differential volume formula for a cylindrical shell (people memorize this):
\[dV ~=~ 2\pi r h\,dr\]
We need to get height as a function of \(r\). We can get this from a difference of the two lines.
\[h = (0.65 r) - (-0.65r)\]
\[h = 1.3 r\]
Now we are ready for the integral. We need to sum up a bunch of differential volumes as \(r\) varies from 2.16 to 9.38.
\[V ~=~ \int_{r=2.16}^{r=9.38}dV\]
For cylindrical shells, \(dV ~=~ 2\pi r h\,dr\), as we proved earlier.
\[V ~=~ \int_{2.16}^{9.38}2\pi r h\,dr\]
Write the expression in terms of \(r\). So substitute \(h = 1.3 r\). Also, constant factors (multipliers) can be taken out from an integral.
\[dV ~=~ \pi r^2 h + 2\pi r h\,dr + \pi h \, (dr)^2-\pi r^2h\]
Notice two terms eliminate because they are opposites.
\[dV ~=~ 2\pi r h\,dr + \pi h \, (dr)^2\]
Remember, \(dr\) is an infinitesimal amount (\(dr\) is near 0), so the second term can be ignored. This leaves us with the differential volume formula for a cylindrical shell (people memorize this):
\[dV ~=~ 2\pi r h\,dr\]
We need to get height as a function of \(r\). We can get this from a difference of the two lines.
\[h = (0.79 r) - (-0.79r)\]
\[h = 1.58 r\]
Now we are ready for the integral. We need to sum up a bunch of differential volumes as \(r\) varies from 2.56 to 7.07.
\[V ~=~ \int_{r=2.56}^{r=7.07}dV\]
For cylindrical shells, \(dV ~=~ 2\pi r h\,dr\), as we proved earlier.
\[V ~=~ \int_{2.56}^{7.07}2\pi r h\,dr\]
Write the expression in terms of \(r\). So substitute \(h = 1.58 r\). Also, constant factors (multipliers) can be taken out from an integral.
\[dV ~=~ \pi r^2 h + 2\pi r h\,dr + \pi h \, (dr)^2-\pi r^2h\]
Notice two terms eliminate because they are opposites.
\[dV ~=~ 2\pi r h\,dr + \pi h \, (dr)^2\]
Remember, \(dr\) is an infinitesimal amount (\(dr\) is near 0), so the second term can be ignored. This leaves us with the differential volume formula for a cylindrical shell (people memorize this):
\[dV ~=~ 2\pi r h\,dr\]
We need to get height as a function of \(r\). We can get this from a difference of the two lines.
\[h = (0.95 r) - (-0.95r)\]
\[h = 1.9 r\]
Now we are ready for the integral. We need to sum up a bunch of differential volumes as \(r\) varies from 2.2 to 8.74.
\[V ~=~ \int_{r=2.2}^{r=8.74}dV\]
For cylindrical shells, \(dV ~=~ 2\pi r h\,dr\), as we proved earlier.
\[V ~=~ \int_{2.2}^{8.74}2\pi r h\,dr\]
Write the expression in terms of \(r\). So substitute \(h = 1.9 r\). Also, constant factors (multipliers) can be taken out from an integral.
\[dV ~=~ \pi r^2 h + 2\pi r h\,dr + \pi h \, (dr)^2-\pi r^2h\]
Notice two terms eliminate because they are opposites.
\[dV ~=~ 2\pi r h\,dr + \pi h \, (dr)^2\]
Remember, \(dr\) is an infinitesimal amount (\(dr\) is near 0), so the second term can be ignored. This leaves us with the differential volume formula for a cylindrical shell (people memorize this):
\[dV ~=~ 2\pi r h\,dr\]
We need to get height as a function of \(r\). We can get this from a difference of the two lines.
\[h = (0.58 r) - (-0.58r)\]
\[h = 1.16 r\]
Now we are ready for the integral. We need to sum up a bunch of differential volumes as \(r\) varies from 6.01 to 9.16.
\[V ~=~ \int_{r=6.01}^{r=9.16}dV\]
For cylindrical shells, \(dV ~=~ 2\pi r h\,dr\), as we proved earlier.
\[V ~=~ \int_{6.01}^{9.16}2\pi r h\,dr\]
Write the expression in terms of \(r\). So substitute \(h = 1.16 r\). Also, constant factors (multipliers) can be taken out from an integral.
\[dV ~=~ \pi r^2 h + 2\pi r h\,dr + \pi h \, (dr)^2-\pi r^2h\]
Notice two terms eliminate because they are opposites.
\[dV ~=~ 2\pi r h\,dr + \pi h \, (dr)^2\]
Remember, \(dr\) is an infinitesimal amount (\(dr\) is near 0), so the second term can be ignored. This leaves us with the differential volume formula for a cylindrical shell (people memorize this):
\[dV ~=~ 2\pi r h\,dr\]
We need to get height as a function of \(r\). We can get this from a difference of the two lines.
\[h = (0.41 r) - (-0.41r)\]
\[h = 0.82 r\]
Now we are ready for the integral. We need to sum up a bunch of differential volumes as \(r\) varies from 2.48 to 6.08.
\[V ~=~ \int_{r=2.48}^{r=6.08}dV\]
For cylindrical shells, \(dV ~=~ 2\pi r h\,dr\), as we proved earlier.
\[V ~=~ \int_{2.48}^{6.08}2\pi r h\,dr\]
Write the expression in terms of \(r\). So substitute \(h = 0.82 r\). Also, constant factors (multipliers) can be taken out from an integral.
A bucket filled with water will be pulled up 34 meters from the bottom of a well. The bucket is pulled up by a heavy rope that connects the pulley to the bucket. As the bucket is lifted, this rope’s length decreases from 34 meters to 0 meters. The bucket and water together have a weight of 40 Newtons. The rope has a linear weight density of 5.8 Newtons per meter.
Each infinitesimal bit of work can be calculated from the product of force times infinitesimal displacement (change in height).
\[dW = F(x) \, dx\]
Notice that the force needed to lift the water, without acceleration, equals the total weight of the bucket, water, and rope between the pulley and the bucket. Because the length of rope decreases, the total weight decreases, so it gets easier to lift the bucket each millimeter.
The total work can be found by adding up all the infinitesimal bits of work. Find the total work in Joules (1 Joule equals 1 Newton\(\cdot\)meter). The tolerance is \(\pm\) 100 J.
Solution
Figure out the force function. The weight of the bucket+water is always 40 Newtons. The rope weight starts at 197.2 Newtons, but decreases to 0 as the bucket is lifted. To find the weight of the rope, you multiply the linear weight density by the length of the rope.
So the work equals approximately 4700 J, or about 1.1 food calories.
Question
A bucket filled with water will be pulled up 14 meters from the bottom of a well. The bucket is pulled up by a heavy rope that connects the pulley to the bucket. As the bucket is lifted, this rope’s length decreases from 14 meters to 0 meters. The bucket and water together have a weight of 41 Newtons. The rope has a linear weight density of 6.8 Newtons per meter.
Each infinitesimal bit of work can be calculated from the product of force times infinitesimal displacement (change in height).
\[dW = F(x) \, dx\]
Notice that the force needed to lift the water, without acceleration, equals the total weight of the bucket, water, and rope between the pulley and the bucket. Because the length of rope decreases, the total weight decreases, so it gets easier to lift the bucket each millimeter.
The total work can be found by adding up all the infinitesimal bits of work. Find the total work in Joules (1 Joule equals 1 Newton\(\cdot\)meter). The tolerance is \(\pm\) 100 J.
Solution
Figure out the force function. The weight of the bucket+water is always 41 Newtons. The rope weight starts at 95.2 Newtons, but decreases to 0 as the bucket is lifted. To find the weight of the rope, you multiply the linear weight density by the length of the rope.
So the work equals approximately 1200 J, or about 0.3 food calories.
Question
A bucket filled with water will be pulled up 54 meters from the bottom of a well. The bucket is pulled up by a heavy rope that connects the pulley to the bucket. As the bucket is lifted, this rope’s length decreases from 54 meters to 0 meters. The bucket and water together have a weight of 77 Newtons. The rope has a linear weight density of 2.8 Newtons per meter.
Each infinitesimal bit of work can be calculated from the product of force times infinitesimal displacement (change in height).
\[dW = F(x) \, dx\]
Notice that the force needed to lift the water, without acceleration, equals the total weight of the bucket, water, and rope between the pulley and the bucket. Because the length of rope decreases, the total weight decreases, so it gets easier to lift the bucket each millimeter.
The total work can be found by adding up all the infinitesimal bits of work. Find the total work in Joules (1 Joule equals 1 Newton\(\cdot\)meter). The tolerance is \(\pm\) 100 J.
Solution
Figure out the force function. The weight of the bucket+water is always 77 Newtons. The rope weight starts at 151.2 Newtons, but decreases to 0 as the bucket is lifted. To find the weight of the rope, you multiply the linear weight density by the length of the rope.
So the work equals approximately 8200 J, or about 2 food calories.
Question
A bucket filled with water will be pulled up 42 meters from the bottom of a well. The bucket is pulled up by a heavy rope that connects the pulley to the bucket. As the bucket is lifted, this rope’s length decreases from 42 meters to 0 meters. The bucket and water together have a weight of 60 Newtons. The rope has a linear weight density of 5.1 Newtons per meter.
Each infinitesimal bit of work can be calculated from the product of force times infinitesimal displacement (change in height).
\[dW = F(x) \, dx\]
Notice that the force needed to lift the water, without acceleration, equals the total weight of the bucket, water, and rope between the pulley and the bucket. Because the length of rope decreases, the total weight decreases, so it gets easier to lift the bucket each millimeter.
The total work can be found by adding up all the infinitesimal bits of work. Find the total work in Joules (1 Joule equals 1 Newton\(\cdot\)meter). The tolerance is \(\pm\) 100 J.
Solution
Figure out the force function. The weight of the bucket+water is always 60 Newtons. The rope weight starts at 214.2 Newtons, but decreases to 0 as the bucket is lifted. To find the weight of the rope, you multiply the linear weight density by the length of the rope.
So the work equals approximately 7000 J, or about 1.7 food calories.
Question
A bucket filled with water will be pulled up 13 meters from the bottom of a well. The bucket is pulled up by a heavy rope that connects the pulley to the bucket. As the bucket is lifted, this rope’s length decreases from 13 meters to 0 meters. The bucket and water together have a weight of 60 Newtons. The rope has a linear weight density of 8.5 Newtons per meter.
Each infinitesimal bit of work can be calculated from the product of force times infinitesimal displacement (change in height).
\[dW = F(x) \, dx\]
Notice that the force needed to lift the water, without acceleration, equals the total weight of the bucket, water, and rope between the pulley and the bucket. Because the length of rope decreases, the total weight decreases, so it gets easier to lift the bucket each millimeter.
The total work can be found by adding up all the infinitesimal bits of work. Find the total work in Joules (1 Joule equals 1 Newton\(\cdot\)meter). The tolerance is \(\pm\) 100 J.
Solution
Figure out the force function. The weight of the bucket+water is always 60 Newtons. The rope weight starts at 110.5 Newtons, but decreases to 0 as the bucket is lifted. To find the weight of the rope, you multiply the linear weight density by the length of the rope.
So the work equals approximately 1500 J, or about 0.36 food calories.
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{1}^{10}f(x)\,dx\]
Solution
The definite integral represents the area under \(y=f(x)\) between \(x=1\) and \(x=10\). Find the area by counting boxes.
The definite integral represents the area under \(y=f(x)\) between \(x=1\) and \(x=10\).
\[\int_{1}^{10}f(x)\,dx = 60.5\]
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{1}^{7}f(x)\,dx\]
Solution
The definite integral represents the area under \(y=f(x)\) between \(x=1\) and \(x=7\). Find the area by counting boxes.
The definite integral represents the area under \(y=f(x)\) between \(x=1\) and \(x=7\).
\[\int_{1}^{7}f(x)\,dx = 16.5\]
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{1}^{5}f(x)\,dx\]
Solution
The definite integral represents the area under \(y=f(x)\) between \(x=1\) and \(x=5\). Find the area by counting boxes.
The definite integral represents the area under \(y=f(x)\) between \(x=1\) and \(x=5\).
\[\int_{1}^{5}f(x)\,dx = 16.5\]
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{0}^{10}f(x)\,dx\]
Solution
The definite integral represents the area under \(y=f(x)\) between \(x=0\) and \(x=10\). Find the area by counting boxes.
The definite integral represents the area under \(y=f(x)\) between \(x=0\) and \(x=10\).
\[\int_{0}^{10}f(x)\,dx = 69\]
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{3}^{8}f(x)\,dx\]
Solution
The definite integral represents the area under \(y=f(x)\) between \(x=3\) and \(x=8\). Find the area by counting boxes.
The definite integral represents the area under \(y=f(x)\) between \(x=3\) and \(x=8\).
\[\int_{3}^{8}f(x)\,dx = 21\]
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{2}^{8}f(x)\,dx\]
Solution
The definite integral represents total of the signed area between the \(x\)-axis and \(y=f(x)\) between \(x=2\) and \(x=8\). Find the area by counting boxes.
The definite integral represents the total signed area of \(y=f(x)\) between \(x=2\) and \(x=8\).
\[\int_{2}^{8}f(x)\,dx = 1.5\]
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{2}^{9}f(x)\,dx\]
Solution
The definite integral represents total of the signed area between the \(x\)-axis and \(y=f(x)\) between \(x=2\) and \(x=9\). Find the area by counting boxes.
The definite integral represents the total signed area of \(y=f(x)\) between \(x=2\) and \(x=9\).
\[\int_{2}^{9}f(x)\,dx = -5.5\]
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{1}^{7}f(x)\,dx\]
Solution
The definite integral represents total of the signed area between the \(x\)-axis and \(y=f(x)\) between \(x=1\) and \(x=7\). Find the area by counting boxes.
The definite integral represents the total signed area of \(y=f(x)\) between \(x=1\) and \(x=7\).
\[\int_{1}^{7}f(x)\,dx = -0.5\]
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{1}^{6}f(x)\,dx\]
Solution
The definite integral represents total of the signed area between the \(x\)-axis and \(y=f(x)\) between \(x=1\) and \(x=6\). Find the area by counting boxes.
The definite integral represents the total signed area of \(y=f(x)\) between \(x=1\) and \(x=6\).
\[\int_{1}^{6}f(x)\,dx = -1\]
Question
A function \(f(x)\) is graphed below.
Evaluate the following integral.
\[\int_{1}^{9}f(x)\,dx\]
Solution
The definite integral represents total of the signed area between the \(x\)-axis and \(y=f(x)\) between \(x=1\) and \(x=9\). Find the area by counting boxes.
The definite integral represents the total signed area of \(y=f(x)\) between \(x=1\) and \(x=9\).
